Question

Solve
$x^2+$ $4x$ $-8$ = 0

Answer 1

Given $x^2+4x-8=0$

$⟹x^2+2\left(x\right)\left(2\right)+2^2-2^2-8=0$

$⟹(x+2{)}^2-4-8=0$

$⟹(x+2{)}^2=12⟹x+2=\pm 2\sqrt{3}⟹x=-2\pm 2\sqrt{3}$